Question: One zero of f(x)= x^3-2x^2-9x+18 is x=2. Find the other zeros of the function.
One xero is x = 2, so one factor is (x - 2). Divide the original cubic by this factor to get the resulting quadratic.
x^3-2x^2-9x+18 = (x - 2)(x^2 - 9) = 0 -- and the difference of two squares factorises as
(x - 2)(x - 3)(x + 3) = 0
Solutions are: x = 2, x = 3, x = -3